Решите систему неравенств. 3x-9< 0 5x+2> 0 заранее !

[tex]\displaystyle 1)\quad f(x)=8x^3-3x^2+4x-2\\\\\text{f}(x)=\int\limits {(8x^3-3x^2+4x-2)} \, dx =\frac{8x^4}4-\frac{3x^3}3+\frac{4x^2}2-2x+\text{c}=\\\\=2x^4-x^3+2x^2-2x+\text{c}, \quad
\text{a}(-1,2)\\\\2=2\cdot(-1))^3+2\cdot(-1)^2-2\cdot(-1)+\text{c}\\\\2=2+1+2+2+\text{c}\\\\\text{c}+5=0\\\\\text{c}=-5\\\\\boxed{\text{f}(x)=2x^4-x^3+2x^2-2x-5}[/tex]

[tex]\displaystyle 2)\quad f(x)=3\cos(3x)\\\\\text{f}(x)=\int\limits {3\cos(3x)} \, dx =\int\limits {\cos(3x)} \,
d(3x)=\sin(3x)+\text{c}, \quad a\bigg(\frac{\pi}{18},\, 1\bigg)\\\\1=\sin\bigg(3\cdot\frac{\pi}{18}\bigg)+\text{c}\\\\\text{c}=1-\sin\bigg(\frac{\pi}6\bigg)=1-\frac{1}2=\frac{1}2\\\\\boxed{\text{f}(x)=\sin(3x)+\frac{1}2}[/tex]

[tex]\displaystyle f(x)=x+2\cos x\\\\f'(x)=1-2\sin x=0\\\\2\sin x=1\\\\\sin x=\frac{1}2\\\\\left[\begin{array}{ccc}\displaystyle x=\frac{\pi}{6}+2\pi n; \quad n\in z\\\\\displaystyle x=\frac{5\pi}6+2\pi n; \quad n \in z\end{array}\right
\\\\\\\underline{-\quad\quad\frac{\pi}6\quad\quad+\quad\quad\frac{5\pi}6\quad\quad-\quad\quad\frac{13\pi}6\quad\quad+\quad\quad\frac{17\pi}6\quad\quad-}[/tex]

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